# Will a warm body naturally slowdown?

Suppose a warm body moving in an empty space with high speed.

The body emits radiation based on its temperature. The protons emitted forwards of the body will have higher energy due to Doppler shift than those emitted backwards. Thus they will care greater momentum. The body should slowdown due to emitted radiation.

Is there a mistake in this reasoning? Possibly the number of photons emitted forwards will be smaller?

• Your reasoning looks fine.. No, the number of photons should stay the same in all directions for a symmetric body. I'm not too sure about this, though; there may be some quantum effects involved. If you're looking at this from the Gallileo point of view (Body in motion in vacuum stays in motion), then the flaw in using this is that there is a net force on the body, as loss of photons $\implies$ change in momentum $\implies$ force (though this comes from Newton). Rockets work on the same principle (they lose mass through their jets). Feb 6, 2012 at 12:59
• The doppler effect is an observer-dependent effect. If this actually happened, it would violate relativity.
– user2963
Feb 6, 2012 at 14:11
• zephyr, indeed but lets consider a rest frame relative to which the warm body moves. Feb 6, 2012 at 14:15
• No, if you are in a frame which is not moving relative to the emitter, they are not shifted.
– user2963
Feb 6, 2012 at 15:10
• No, it seems it will not slow down even though it will loose the momentum. It will have lesser momentum and lesser mass but the same speed. Feb 6, 2012 at 15:34

The apparent paradox can be analyzed more clearly if we simplify it a bit. Let's assume we have one object with mass $M$ emitting two photons in opposite directions, each of them with momentum $p$.

In the rest frame of the object, choosing the X-axis to align with the emitted photons, we will have the following four-momenta (using units where $c = 1$):

Before

$p_{obj}^{before} = [M,0,0,0]$

After

$p_{obj}^{after} = [M-2p,0,0,0]$

$p_{\gamma_1} = [p,p,0,0]$

$p_{\gamma_2} = [p,-p,0,0]$

Here we clearly have four-momentum conservation:

$p_{obj}^{before} = p_{obj}^{after} + p_{\gamma_1} + p_{\gamma_2}$

As this is a Lorentz-invariant four-vector equation, it will be valid in any reference frame.

To see it more explicitly, let's boost the system by a speed $v$ in the +X direction and see how the equations look then:

Before

$p_{obj}^{before} = \left[M\,(1-v^2)^{-1/2},M\,v\,(1-v^2)^{-1/2},0,0\right]$

After

$p_{obj}^{after} = \left[(M-2p)\,(1-v^2)^{-1/2},(M-2p)\,v\,(1-v^2)^{-1/2},0,0\right]$

$p_{\gamma_1} = \left[p(1-v^2)^{-1/2}+p\,v\,(1-v^2)^{-1/2},p(1-v^2)^{-1/2}+p\,v\,(1-v^2)^{-1/2},0,0\right]$

$p_{\gamma_1} = \left[p(1+v)(1-v^2)^{-1/2},p(1+v)(1-v^2)^{-1/2},0,0\right]$

$p_{\gamma_2} = \left[p(1-v)(1-v^2)^{-1/2},p(1-v)(1-v^2)^{-1/2},0,0\right]$

Here it's a bit more tedious, though straightforward, to check four-momentum conservation.

To finish the analysis, let's see how the X component of the photon momentum is transformed when we boost the system. We start by adding back the powers of $c$:

$p^{boosted}_{\gamma_1x} = p\left(1+\frac{v}{c}\right)\left(1-\frac{v^2}{c^2}\right)^{-1/2}$

Using the relationship between wavelength and momentum and doing some algebra,

$\lambda^{boosted}_{\gamma_1} = h\left(\frac{h}{\lambda^{unboosted}_{\gamma_1}}\left(1+\frac{v}{c}\right)\left(1-\frac{v^2}{c^2}\right)^{-1/2}\right)^{-1}$

$\lambda^{boosted}_{\gamma_1} = \lambda^{unboosted}_{\gamma_1}\left(1+\beta\right)^{-1}\left(1-\beta^2\right)^{1/2}$

$\lambda^{boosted}_{\gamma_1} = \lambda^{unboosted}_{\gamma_1}\left(1+\beta\right)^{-1}\left(1+\beta\right)^{1/2}\left(1-\beta\right)^{1/2}$

$\displaystyle \frac{\lambda^{boosted}_{\gamma_1}}{\lambda^{unboosted}_{\gamma_1}} = \sqrt{\frac{1-\beta}{1+\beta}}$

we get the expression of the wavelength change by the relativistic Doppler effect (the difference in the sign of $\beta$ is due to the velocity sign convention).

Well the answer is that the body will indeed loose the momentum. But since the mass of the body will decrease as well due to radiation, the velocity should not change.

• Uhh, how exactly will the mass decrease? A body that cools down does not lose mass (in the classical sense). The photons are created by a decrease in thermal energy of the body. Feb 6, 2012 at 15:51
• Loosing energy it will decrease its total inertial mass. Because thermal energy has mass of its own. The total momentum of energy (thermal+radiation) will not change. Feb 6, 2012 at 16:02
• Thats why I said 'in the classical sense' im neglecting relativity. Oh! I got the solution now... I'll post it as an answer.. Feb 6, 2012 at 16:08
• Relativity should not be neglected here because the radiation carries some momentum. If to conut the momentum carried by radiation, one should also count the momentum carried by the thermal energy inside the body. Feb 6, 2012 at 16:12
• The momentum carried by radiation is a quantum mechanical effect, not a relativistic effect. But you're right, we cannot neglect relativity. I'm posting it as an answer (will take 10 mins, not doing this from a pc), as its a bit long. Feb 6, 2012 at 16:19

The body will not slow down - that would contradict relativity. While the following phrase in the answer: "The body emits radiation based on its temperature" is technically correct, that does not mean that the radiation energy distribution is the same for a moving body and a body at rest. For the moving body you should use electrodynamics of moving bodies.

(This is a qualitative approach only, but I'll give enough data so that you can solve it for yourself) Alright. The main issue here is that we are neglecting relativisticc effects, which HAVE to be taken into account while dealing with the doppler effect of light.

In the doppler effect of sound, we always assume the medium as a 'rest frame', as sound has a fixed speed relative to its medium. The doppler effect comes partially from this fact. Now, as light has a fixed vacuum speed, in this situation, a rest frame has no meaning. There still is a doppler effect, it is known as the relativistic doppler effect. As with all relativistic things, this is only significant at large velocities. But the paradox is resolved at whatever velocity you take, as long as we do not neglect this and one more thing:

The momentum of a body in relativity is $\gamma m_0 v$, where $\gamma$ is the lorentz factor, and $m_0$ is the rest mass. Note that the rest mass is the total energy of the body at rest divided by $c^2$. Now, as the body loses thermal energy (it's cooling), this value of $m_0$ also decreases. Thus $\gamma m_0$ decreases (by a small amount, unless v is large)

Now, the conjunction of the two gives us this: The momentum of the body decreases, but so does its mass. This should hopefully lead to no net slowing down. The reason for it not slowing down was well given by @zephyr above. I'll just elaborate it here: Consider a different reference frame, moving with velocity 2v forwards. Now, in this frame, the situation is identical except that the body is moving backwards. Let us assume that it slows down in our initial frame. As all inertial frames are equivalent, it should slow down relative to the new frame, too. But, this slowing down will be in an opposite direction, and thus the two observers will disagree on the direction of the body's deceleration. This is not permitted in inertial frames. So by reducto ad absurdum, the initial premise(the body decelerates) is false. To prove it another way, Just draw a diagram with the velocities and relative velocities and you'll get $-a=a\implies a=0$

Of course, you can quantitatively calculate the change in rest mass and momentum and (hopefully) prove that the the velocity does not change.

Lets see.

A photon carries off momentum $h\nu/c$.

Now if the body is a round body, a star lets say, it will be radiating uniformly from the surface. Lets take the forward going ones, their $\nu$ increases due to the forward motion, and the momentum increases. If we take the backward going ones the $\nu$ decreases and the momentum decreases. One would have to compute it, I would say that the forward and backward going momenta will add up to zero change for the body, at first order since they are linear with the wavelength changes, thus not changing the velocity of the body. Calculating for a particular shape and a particular radiation spectrum might give higher order effects .

• If absolute value of the momentum of the photons going forwards is greater than that of photons going backwards, how can they sum up to zero? Feb 6, 2012 at 13:27
• I guess I am saying that a photon of momentum p leaving in front gains from the motion a +delta(p) and a photon of momentum p leaving behind gets a -delta(p) so the momentum of the body is not affected by the radiation of these two photons. Feb 6, 2012 at 13:39
• I think that the problem comes from ignoring the body. The forward going photons have more momentum and the backward going ones less to the amount the forward took off, leaving the body at the same momentum. I will edit the answer. Feb 6, 2012 at 13:51
• Due to Doppler effect their moments cannot be equal in absolute value. Feb 6, 2012 at 14:03
• Those going forward gain in the absolute value of momentum and those going backaward loose in absolute value of momentum, but the both deltas directed to the same direction (forward), so they both effectively slow down the body's motion. Write the formulas down to see yourself. You forgot that momentum is a vector value, not scalar like kinetic energy. Feb 6, 2012 at 14:19