homework and exercises - Proof of centripetal acceleration formula ($a_c = v^2/r$) for non-uniform circular motion

The formula for centripetal (radial) acceleration is well known, and there exist many proofs for it: $$||a_c|| = \frac{||v||^2}{r}$$

However, all the proofs I've seen rely on the fact that it is uniform circular motion and the magnitude of the tangential velocity vector does not change. For instance, take the classic proof using similar triangles — the similarity can only be established if the final tangential velocity vector and the initial one are of the same length.

In addition, take this calculus based proof from Khan Academy, outlined as follows:

For this proof to work, $(d\theta / dt)$ must be considered a constant, $\omega$, that does not depend on time. In the case of non-uniform circular motion, however, this is not always true as since there exists a tangential acceleration along with a radial one, $\omega$ must depend on time and is not necessarily a constant value.

Intuitively I understand that the centripetal/radial acceleration depends only on the difference in orientation between two tangential velocity vectors, and that their magnitudes do not matter -- hence the formula intuitively holds true in the non-uniform case. However, how would you go about modifying either of the proofs presented so that they are still valid in this case? Or alternatively, is there another proof that holds valid even when there exists tangential acceleration?

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As per the suggestion of one of the answers, I let $\omega$ vary with time and took its derivative as $\alpha(t)$. This is my work so far. Unfortunately, I am stuck after the last step.

\begin{align} \overset{\rightharpoonup }{p}(t) &= r \cos (\theta (t))\cdot\hat{i}+r \sin (\theta (t))\cdot\hat{j} \\ \overset{\rightharpoonup }{v}(t) &= -r \sin (\theta (t))\cdot\omega(t)\cdot\hat{i}+r \cos (\theta (t))\cdot\omega(t)\cdot\hat{j} \\ \overset{\rightharpoonup }{a}(t) &= (-r \cos(\theta(t))\cdot\omega(t)^2 - r \sin(\theta(t))\cdot\alpha(t))\hat{i} \\ &+ (-r \sin(\theta(t))\cdot\omega(t)^2 + r \cos(\theta(t))\cdot\alpha(t))\hat{j} \end{align}

From here on, $\theta(t)$ is represented as just $\theta$ for brevity and clarity

$$\overset{\rightharpoonup }{a}(t) = -\omega(t)^2(r \cos\theta\cdot\hat{i} + r \sin\theta\cdot\hat{j}) -\alpha(t)(r \sin\theta\cdot\hat{i} - r \cos\theta\cdot\hat{j})$$

Answer

The proper derivation of the centripetal acceleration—without assuming any kinematic variables are constant—requires a solid understanding of both the stationary Cartesian unit vectors $\hat{i}$ and $\hat{j}$ as well as the rotating polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$. The Cartesian unit vectors $\hat{i}$ and $\hat{j}$ are stationary and always aligned with the X and Y axes respectively, while the polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$ rotate with an angular velocity of $\omega=\|\dot{\theta}\|$ and point in the directions of increasing radius and angle (respectively). The included graphic below shows the two basis vector pairs overlaid on top of one another.

The position vector of the object is obviously defined as:

$\vec{p}(t)=x\hat{i}+y\hat{j}=rcos(\theta)\hat{i}+rsin(\theta)\hat{j}$,

with

$\|\vec{p}(t)\|=\sqrt{(rcos{\theta})^2+(rsin{\theta})^2}=\sqrt{r^2(sin^2(\theta)+cos^2(\theta))}=r\sqrt{(1)}=r$

Less obviously, it can be shown that the polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$ can be expressed solely in terms of the Cartesian unit vectors $\hat{i}$ and $\hat{j}$ and the angular position $\theta$ as,

$\boxed{\hat{e}_r=cos(\theta)\hat{i}+sin(\theta)\hat{j}}$ and $\boxed{\hat{e}_\theta=-sin(\theta)\hat{i}+cos(\theta)\hat{j}}$.

These two equations are extremely important, as they will be the key to expressing the Cartesian acceleration in polar coordinates, of which one of the terms will be our desired $v^2/r=\omega^2r$ centripetal acceleration. Moving forward, the vector acceleration of the object in Cartesian coordinates is simply

$\vec{a}(t)=\frac{d^2}{dt^2}\left[\vec{p}(t)\right]=\ddot{x}\hat{i}+\ddot{y}\hat{j}$.

Starting with $x=rcos(\theta)$ and $y=rsin(\theta)$ and differentiating once, we have

$\boxed{\dot{x}=\dot{r}cos(\theta)-r\dot{\theta}sin(\theta)}$ and $\boxed{\dot{y}=\dot{r}sin(\theta)+r\dot{\theta}cos(\theta)}$.

Differentiating again, we will have

$\ddot{x}=\ddot{r}cos(\theta)-\dot{r}\dot{\theta}sin(\theta)-\dot{r}\dot{\theta}sin(\theta)-r\frac{d}{dt}\left[\dot{\theta}sin(\theta)\right]$

$=\ddot{r}cos(\theta)-2\dot{r}\dot{\theta}sin(\theta)-r\left[\ddot{\theta}sin(\theta)+{\dot{\theta}}^2cos(\theta)\right]$, such that

$\boxed{\ddot{x}=(\ddot{r}-r\dot{\theta}^2)cos(\theta)+(r\ddot{\theta}+2\dot{r}\dot{\theta})(-sin(\theta))}$.

Similarly, the y acceleration $\ddot{y}$ becomes

$\ddot{y}=\ddot{r}sin(\theta)+\dot{r}\dot{\theta}cos(\theta)+\dot{r}\dot{\theta}cos(\theta)+r\frac{d}{dt}\left[\dot{\theta}cos(\theta)\right]$

$=\ddot{r}sin(\theta)+2\dot{r}\dot{\theta}cos(\theta)+r\left[\ddot{\theta}cos(\theta)-{\dot{\theta}}^2sin(\theta)\right]$, such that

$\boxed{\ddot{y}=(\ddot{r}-r\dot{\theta}^2)sin(\theta)+(r\ddot{\theta}+2\dot{r}\dot{\theta})cos(\theta)}$.

Now, we must plug these scalar derivatives into our formulation for the vector acceleration. In Cartesian coordinates, this is

$\vec{a}(t)=\ddot{x}\hat{i}+\ddot{y}\hat{j}=\{(\ddot{r}-r\dot{\theta}^2)cos(\theta)+(r\ddot{\theta}+2\dot{r}\dot{\theta})(-sin(\theta))\}\hat{i}+\{(\ddot{r}-r\dot{\theta}^2)sin(\theta)+(r\ddot{\theta}+2\dot{r}\dot{\theta})(cos(\theta))\}\hat{j}$

which can be rearranged into the following form:

$\vec{a}(t)=(\ddot{r}-r\dot{\theta}^2)\{cos(\theta)\hat{i}+sin(\theta)\hat{j}\}+(r\ddot{\theta}+2\dot{r}\dot{\theta})\{-sin(\theta)\hat{i}+cos(\theta)\hat{j}\}$

But as we have already seen, this is simply equal to

$\boxed{\boxed{\vec{a}(t)=(\ddot{r}-r\dot{\theta}^2)\hat{e}_r+(r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{e}_\theta}}$

As we can now appreciate from carrying out the full derivation, there are actually two components each to both the radial and tangential accelerations. The $\ddot{r}$ term is straightforwardly equal to the second derivative of the position vector magnitude. The second term, $r\dot{\theta}^2$, is our long sought-after centripetal acceleration $r\dot{\theta}^2=\omega^2r=v^2/r$, and (as expected) it points in the negative radial direction. The tangential terms are perhaps a bit less intuitive. The $r\ddot{\theta}$ term is the acceleration that occurs whenever the radius and angular acceleration $\ddot{\theta}$ are both non-zero (imagine the tangential acceleration of a turbine blade of a jet engine as the engine spools up). The final term $2\dot{r}\dot{\theta}$ is what's commonly known as the Coriolis acceleration, and it occurs whenever the radius and angle change simultaneously. It arises because, for a given angular velocity, the arc length travelled every second increases with radius (tangential velocity increases with radius). Thus, an object with a given angular velocity will have different tangential velocities at different local radii of rotation. If the radius changes with time ($\dot{r}\not=0$) and the angular velocity $\dot{\theta}$ is not equal to zero, then the tangential velocity will change with time, which is by definition a tangential acceleration.

energy - Can change in temperature cause a change in mass of an object?

If a gold bar is heated to say 200 degree Celsius then will it have the same mass at say 10 degree Celsius. Does energy has mass? If so then does this increased 'heat energy' cause an increase in the mass of an object

Answer

A gold brick is made of gold atoms that mutually interact. At a given temperature it weighs bit less than the weight of each piece. At a higher temperature it weighs a bit more than at a colder temperature. So the weight isn't the sum of the weights of the parts, not quite.

Even in a single gold atom, it weighs a little bit less than weight of each neutron and proton and electron all added up. And even for a lone proton or neutron the weight isn't the sum of the weights of the quarks inside.

Why? Well, firstly weight isn't caused by mass, it is caused by energy and momentum and pressure and stress. But there is energy associated with mass and for many systems that source, the energy associated with mass is the largest, so large by far that you get almost the correct answer by adding up just that.

But there are other meanings of mass. Does a big chunck of gold resist forces more when it is hotter? Yes it does.

The way you resist forces is related to how you balance energy and momentum. There is energy associated with mass and there is energy associated with momentum. For a single particle $E^2=c^2\vec p^2 + m^2c^4$. So at first when you add a bit of momentum the energy increases by about $ \vec p^2 /2m$ but later after you've added lots of energy from momentum a little bit of additional momentum $\Delta \vec p$ adds about $\vert c\Delta \vec p \vert$ of additional energy. OK, so for a single particle the mass is about how you balance energy and momentum.

Now, if you have a system with a total system energy of E and a total system momentum of $ \vec p$ then the system can act like a giant particle of mass $\sqrt{E^2-c^2\vec p^2}/c^2$. And now you can see that heating the gold brick in the frame where it has no total momentum doesn't increase the total momentum (heating gives more momentum equally in all directions so the total momentum stays zero) so it increases the total system energy and so the system starts to act like a giant particle with a larger mass.

Since that balance of energy and momentum is what determines the kinematic and dynamical aspects of mass like how hard you have to push to get it to go faster the hot brick has a bigger mass.

So by every possible sense. It has more mass. It's still a composite object so the mass of the system (the brick) is not the sum of the masses of the parts. But we've learned that that never happens anyway so we can deal.

So wait you might say. If the mass or weight is never actually the perfect sum of the masses or weights. At least why does it appear to be the sum?

Good question. Imagine that for each particle you had a vector in spacetime with components $(E/c, p_x, p_y, p_z)$ its like a 4d momentum. It turns out that it points in the 4d direction the particle travels in spacetime, and for two particles moving the same way in space time (so traveling from point 1 at time 1 to place 2 at time 2) if one has twice the mass it has twice the 4d momentum. So in a sense the mass is the length of the 4d momentum and the direction is the direction in spacetime the particle is going.

But if you have a bunch of particles that are all moving slow ... relative to each other. Then in a geometric amount of time they all moved almost the same way. So in fixed time interval they all ended up at really close places, so those directions in spacetime are all pointing in almost the same direction.

So their masses are the lengths of their 4d momentums and the 4d momentums are pointing in almost the same directions.

Here is an insight from geometry. The sum of vectors has a length that is almost exactly the sum of the lengths when they are all pointing in almost the same direction. If they are moving slowly relative to each other then those 4d vectors are pointing in almost the same direction so he length of the sum is close to the sum of the lengths. And if you sum those vectors you get the total energy and the total momentum.

So the mass is really close to the sum of the masses when everything is moving slow relative to each other. And in our daily experience that is what happens even a performance race car or high speed jet is slow compared to light. And when we compared the size of that fixed time interval to the different places of the particles we compared the ratio to the speed of light so for everyday life it was a small difference.

There is one difference between usual length and the mass as length of a vector idea I showed. The mass is $\sqrt{E^2-c^2\vec p^2}/c^2$ which has a minus sign. But it is like a length and the geometric fact that the sum of vectors has a length that is almost exactly the sum of the lengths when they are all pointing in almost the same direction, it still holds when you use this kind of a way to measure lengths.

So, yes a hot brick weighs more because of the increased energy. And it resists motion more because you increased the energy by distributing the additional momentum so that in the frame where the total momentum was zero it continues having a total momentum of zero, so the energy in that frame could go to increasing the mass of the system (the length of the 4d momentum of the system).

Local Charts in General Relativity

We may consider a "local" region in curved spacetime (local in respect of the spatial and the temporal coordinates). A "local inertial frame" may be constructed by some transformation that produces flat spacetime locally. This transformation produces the diagonal [1,-1,-1,-1] in an approximate manner.

A physical point:

You are performing some experiment in a small laboratory room (stationary w.r.t. the Earth) for a small period of time. Are you in a local inertial frame? If there was a freely falling lift in front of you, it would have been a better approximation to the inertial frame concept. So your laboratory does not qualify to be an inertial frame when compared with the falling lift. (The difference becomes even more conspicuous if you imagine "gravity" to be a 100 or a thousand times stronger.)

Now, let's move into the problem:

We take a small region of curved spacetime and use some suitable transformation to produce a "local inertial" frame. For this transformation we are disregarding all the anisotropies and inhomogeneities of the surrounding space in the original manifold.Incidentally the objective behind creating local inertial frames is to apply SR. The associated Lorentz transformations are supposed to hold good in the ideal conditions of isotropy and homogeneity of space

Our transformation transforms a small region of curved spacetime to a small region of flat spacetime. To this small/finite flat spacetime, we add the rest of it, ignoring all the anisotropies and inhomogeneities of the surroundings in the original curved spacetime [original manifold]. In effect we are assuming a "Global Transformation" without working it out! (And we expect correct results when we go back to the original space by the reverse transformations) We are in effect assuming a "Global Transformation" from curved spacetime to flat spacetime. Is it not as sacrilegious as "Flattening a Sphere"?

How much unreliability is the aforesaid error going to cast on calculations performed in the local inertial frames?To what extent will this error affect the Principle of Equivalence?

homework and exercises - My book uses helium weight in reaction formula instead of weight alpha particle?

I'm suppose to write out reactions where atoms send out alpha radiation and decay. The book uses the 4-2 H, 4 as nucleon number and 2 as proton number, but isn't that wrong? The mass of helium is greater than the alpha particle due to two electrons? Shouldn't they use a different notation for the alpha-particle other than that for helium?

quantum field theory - How to describe time evolution in relativistic QFT?

I must confess that I'm still confused about the question of time evolution in relativistic quantum field theory (RQFT). From symmetry arguments, from the representation of the Poincare group through unitary operators on a Hilbert space, one knows that the generator of time translations is the Hamiltonian operator ($H$) of the field, and the unitary operator corresponding to a finite time translation from $t_0$ to $t_1$ is $U = \exp[-iH(t_1 - t_0)]$ (with $\hbar = 1$). So apparently, from symmetry arguments, the Schrodinger picture is live and well (i.e., it's valid and must work) in RQFT. However, I've never seen in the literature on RQFT a state vector describing the state of the quantum field at a finite time $t$, that is, I've never seen a $|\psi (t)\rangle$. Then, there is the paper by Dirac in which he takes issue with the Schrodinger time-evolution operator $U = \exp[-iH(t_1 - t_0)]$, practically showing that it doesn't make any sensible sense ("Quantum Electrodynamics without dead wood" published in Phys. Rev., http://journals.aps.org/pr/abstract/10.1103/PhysRev.139.B684).

I would very much appreciate it if you could let me know if $U = \exp[-iH(t_1 - t_0)]$ is indeed the time evolution operator in RQFT, and if the answer is yes, then why the state vector $|\psi (t)\rangle$ is never defined and calculated for a finite time $t$. This question is also important for the path-integral quantization, since in order to construct a path integral one uses the expression $U = \exp[-iH(t_1 - t_0)]$ for the time evolution operator, and also state vectors at finite time. Otherwise, the path integral approach cannot be constructed and the path-integral formulas must be postulated! Are they merely postulated?

EDIT 1: There seems to be a lot of confusion in the comments and answers between the Schrodinger picture and the Schrodinger representation (in QFT). As in all the textbooks on QM there are, by the Schrodinger picture I understand a description of a quantum system by states represented by abstract time-dependent state vectors in some Hilbert space, and whose observables are represented by time-independent operators on that Hilbert space. There is nothing unclear here, and there is no question of any ill-defined "functionals" since no representation has been introduced yet, i.e., no particular basis of the Hilbert space has been chosen in order to pass from abstract vectors to wave functions (or wave functionals) by projecting the abstract vectors on that basis.

As a matter of fact, Dirac in his paper uses abstract vectors in the formalism of second quantization. What is really puzzling, and this is the main cause of my confusion, is that the time-evolution operator should exist in the Schrodinger picture (i.e., there must be a unitary operator relating any two state vectors that have different time arguments), based on the Poincare invariance imposed on any relativistic quantum system. Yet, as Dirac shows the math just doesn't make any sense (and this is not because of some "functionals") and it seems that the Schrodinger picture must be banned from QFT, and this, in turn, comes against Poincare invariance!

EDIT 2: It's interesting to note that $U(t)$ doesn't exist even in the Heisenberg picture and not even for a free field. V. Moretti's construction is faulty since his $U(t)$ doesn't have a domain of definition. Indeed, $ \psi(x) \propto a_p + a_p^{\dagger}$. Therefore, $(\psi(x))^2 \propto (a_{p})^2 + (a_p^{\dagger})^2 + ...$, and hence $H \propto \int (\psi(x))^2 d^3x \propto \sum_p \{(a_{p})^2 + (a_p^{\dagger})^2 + ...\}.$ Hence, $U(t) \propto t^2 \sum_p (a_{p})^{2}(a_p^{\dagger})^{2}$ already at second order in $t$. Hence, acting on the vacum state $U(t)$ gives $\infty$ (for finite $t$!), as Dirac has shown in his paper. Therefore, $U(t)$ doesn't have a domain of definition and it cannot exist, even in the Heisenberg picture!

EDIT 3: @Valter Moretti You claim in your answer that you work entirely in the Heisenberg picture. Therefore, the field depends on both space $\bf{x}$ and time $t$. However, when you construct the Hamiltonian you "smear" the field with a test function, but integrate only over the space, i.e. over $\bf{x}$, but not over time $t$. Therefore, your Hamiltonian operator should depend on time $t$. You cannot simply eliminate $t$ unless you also integrate over it! There is no such construction in the literature! If it's not so, please edit your answer by showing as to how you get rid of the time dependence in the Hamiltonian.

astronomy - About the hump on galaxy rotation curves

The past days I have been studying the rotation curves of disk galaxies and I am currently trying to understand how we can extract information about the dark matter of a galaxy by looking its rotation curve. A typical and well behaved rotation curve of a disk galaxy rises fast at the start, then it peaks and produces a "hump"/"bump" and then it remains constant (or slightly declining). I would like to ask what exactly causes the "hump"/"bump" on the rotation curve and what it can tell us about the galaxy.

Answer

The first hump that you see is caused by the galactic disk. The gravitational potential for this is given by* $$ \Phi_{disk}=-\frac{GM_{disk}}{\left(r^2+(a+\sqrt{b^2+z^2})^2\right)^{1/2}} $$ where $a$ is the scale radius of the disk, $b$ the scale height, and all other variables take their usual meaning. The rotational velocity is then the square root of the $r$-derivative of $\Phi$ times r: $$ v_{rot,disk}=\left(\frac{GM_{disk}r^2}{\left(r^2+(a+\sqrt{b^2+z^2})^2\right)^{3/2}}\right)^{1/2} $$ If you plot this (with the appropriate units), you should get that first bump you ask about.

image source

This particular bump can tell us two things:

1. The scale radius


2. The mass of the disk

A smaller value of $a$ will lead to a peak at shorter $r$ while also increasing the rotational velocity; a larger $a$ will lead to the opposite change. A larger $M_{disk}$ will produce a larger peak rotational velocity (assuming you fixed $a$, your model's peak would occur at the same point).

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*There are other models of potentials, see Section 7.4 of this link; the model I am using is called the Miyamoto-Nagai potential.

optics - Why can colors be mixed?

We can combine colored light, creating other colors, at least in terms of visual perception.

But how it the result physically "a different color" - if it is at all?

Or is all this not a physical question to begin with - but only about our eye and brain?

To have an example, we

• have an incandescent bulb, showing "white" light, and


• combine red, green and blue light in intensities such that it looks roughly the same.

It is not central to the question whether it is exactly matching the white light - but certainly interesting to understan whether it could perfectly match, and why.

Answer

A "normal, healthy" human eye has two types of light-sensitive cells in the retina: rods ("color blind", but capable of sensing low light levels) and cones: cells that are sensitive to different bands. See this figure for their relative sensitivity (from http://hyperphysics.phy-astr.gsu.edu/hbase/vision/colcon.html)

When you look at a spectrum of light, you can think of the eye's response as the result of the integral of $\int{I(\lambda)* sensitivity(\lambda)*d\lambda}$ for each of the three types of cones. It should be immediately obvious that different spectra (i.e. different "colors") can be perceived as being "the same" because of this way of processing the light. It gets further complicated because some people may be missing one or more types of cones ("color blind") so their perception of some colors may be different altogether.

The above is the reason that making "good white light" is hard. If you start with a bright (hot) incandescent source, you will be illuminating an object with every wavelength in the visible spectrum; but fluorescent lights (and LEDs) tend to create light in "bands". Thus, your perception of some colors will be different depending on the light source used. This is usually reflected in something called the "color rendering index" of the light source - where "100" represents perfect rendering, and lower numbers are indicative of a "lumpy" spectrum. The eye might still see this as "white" (because the source light excites the three types of cones evenly), but other "colors" may not show up very well since they do not in fact get illuminated.

An example of the different spectra is given at http://www.ledsmagazine.com/content/dam/leds/migrated/objects/features/10/2/11/Avnet_Fig2T_22513.jpg

You can see that a typical incandescent lamp is biased towards red (because the filament is not as hot as the sun - this is why digital cameras do "white balance correction" and they usually have a "tungsten filament" setting). By contrast, a fluorescent lamp tends to have several distinct peaks in the emission spectrum - these are the emission peaks of the phosphors used to convert the UV light emitted in the gas of the tube into visible light. Finally, modern white LED technology is reaching a point where quite a wide range of excitation wavelengths are present, leading to higher color rendering fidelity.

Looking at the spectrum of the fluorescent tube above, if you had an object that was "black" except for a reflection in a narrow band at 570 nm, it would look black with such a source, when it ought to look orange. Yet if the same object was reflective at 545 nm (the emission peak), it would look bright green. An object that was a mix of "colors" in that narrow range would look very different in fluorescent light than in sunlight.

So the answer to your question is really - it's in the eye, not the brain.