Does Light Obey the Same Rules as Matter in Motion?

[dayalanand roy]

Does light have inertia?
Suppose a light pulse is emitted in a moving train in the direction of train’s movement. Should it move at its constant speed even in the moving train? And if there is a front window in the path of the light pulse, should it shoot out through the window to go ahead of the train and leave the train to lag behind? When we walk in the train in the direction of train’s movement, to an observer on the ground, our velocity will be added to train’s velocity. But should this be the case with the light pulse which will never show zero velocity in the train even to an observer aboard the train, and hence, should its velocity be added to train’s velocity to an observer on the ground. Now suppose that the train is moving at a speed much faster than light. The passengers sitting inside will also be moving at faster than light speed to observers on the ground. But should not the light pulse again travel at its own constant speed inside the train. Now if there is a rear window in the train, should the light pulse be, after some time, thrown back from the train through this window to lag behind the train, because it has little or no inertia? So, I want to know, whether the light pulse should face velocity addition problem or not? If it does not show velocity addition problem, can we use the speed of light as a reference for time? If we can, will it not lead us back to the absoluteness of time?

 

[jtbell]

Suppose a light pulse is emitted in a moving train in the direction of train’s movement. Should it move at its constant speed even in the moving train?

Yes.

And if there is a front window in the path of the light pulse, should it shoot out through the window to go ahead of the train and leave the train to lag behind?

Yes.

When we walk in the train in the direction of train’s movement, to an observer on the ground, our velocity will be added to train’s velocity.

Not in the way you're probably thinking. Look up "relativistic velocity addition", for example here:

http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/relativ/einvel.html

For everyday speeds, the result is very very close to simple addition. When the speeds are a significant fraction of c, the result is significantly different from simple addition.

But should this be the case with the light pulse

The formula described at the link above, also works with light, but it always gives c for the velocity of the light pulse, no matter what speed (less than c) you try to "add" it to.

Now suppose that the train is moving at a speed much faster than light.

Sorry, no can do.

 

[Dale]

Does light have inertia?

A single pulse of light has energy and momentum, but not mass. A system of multiple pulses of light can have mass also. So I would say yes.

 

[dayalanand roy]

Many thanks for the reply.
Regards

 

[dayalanand roy]

Thanks for the reply.
regards

 

[smyth]

What if I shoot a photon upwards in that train? Will the direction of the photon be altered by the speed of the train?

 

[WannabeNewton]

What if I shoot a photon upwards in that train? Will the direction of the photon be altered by the speed of the train?

Relative to a person on the ground yes. Relative to you, no.

 

[dauto]

What if I shoot a photon upwards in that train? Will the direction of the photon be altered by the speed of the train?

Yes, that change of the direction of propagation of light is called stellar aberration and must be taken into account by precise astronomical observations due to the motion of the Earth around the sun.

 

[AdrianLuther]

A single pulse of light has energy and momentum, but not mass. A system of multiple pulses of light can have mass also. So I would say yes.

Can you explain how multiple pulses of light can have mass?

 

[xox]

Can you explain how multiple pulses of light can have mass?

The energy and momentum of the photon are related via E=pc.
The mass of a system of particles satisfies the relationship: (Mc^2)^2=E^2-(\vec{p}c)^2.
1. For one photon, the above becomes: (Mc^2)^2=(pc)^2-(\vec{p}c)^2=0
2. For two photons with identical momenta: (Mc^2)^2=(2pc)^2-(2\vec{p}c)^2=0

But...

3. For two photons with opposing momenta: (Mc^2)^2=(2pc)^2-(0)^2=(2pc)^2 \ne 0

The above generalizes to the case of multiple photons:

(Mc^2)^2=(\Sigma{pc})^2-(c \Sigma{\vec{p}})^2=c^2((\Sigma{p})^2-(\Sigma{\vec{p}})^2)

In general, (\Sigma{p})^2 \ne (\Sigma{\vec{p}})^2 so, M>0.

PS: It is easy to prove that (\Sigma{p})^2 \ge (\Sigma{\vec{p}})^2