How to understand why massless particles must travel at c?

[SecretSnow]

First of all, i need to say that i have read various threads posted by other users in physicsforum who have asked the same question. However, I still don't have a clear picture of why massless particles must travel at c. Personally, I have came up with an explanation which is rather simple, which may or may not be valid, and of course, i wish to validate it here. Another thing I wish to know is why must massless particles travel at c according to the 'equation way'? That is, through this equation $E^2= (mc^2)^2 + (pc)^2$

This is my own guess: All objects with mass can reach c, given that there is time for it to accelerate all the way to c. However, for massless particles they reach c immediately because they do not need time to accelerate and they reach c in no time, or 0 seconds. Hence, we perceive that all massless particles must travel at c, since there is no way in which we can view a massless particle to be traveling at a speed lower than c because it just reaches c instantly. Is this reasoning valid? However, there's another question. Am i right to say that as long an object is allowed to accelerate indefinitely in space in the same direction without any resistance or collision, it will reach the cosmic speed limit?

Of course, I would definitely like to understand this concept through the 'legitimate' way or 'equation way'. These are some links which otaher users have posted about the same question:

https://www.physicsforums.com/showthread.php?t=502572
https://www.physicsforums.com/showthread.php?t=202934
https://www.physicsforums.com/showthread.php?t=481362
https://www.physicsforums.com/showthread.php?t=396044

 

[mfb]

All objects with mass can reach c

No.

However, for massless particles they reach c immediately because they do not need time to accelerate

They do not even accelerate.$E^2= (mc^2)^2 + (pc)^2$
$E=pc \frac{c}{v}$
For massless particles, E=pc and therefore v=c.

 

[jtbell]

A useful general relationship for the velocity as a fraction of the speed of light is

$$\frac{v}{c} = \frac{pc}{E}$$

For example, an electron with momentum 300 keV/c and energy 400 keV has v/c = 0.75, i.e. v = 0.75c.

For a photon, pc = E.

This is my own guess: All objects with mass can reach c, given that there is time for it to accelerate all the way to c.

No. There is never enough time to reach c, even in principle. It would require delivering an infinite amount of energy to the object, which would require an infinite amount of time. You can get it infinitesimally close to c; how close depends on how much energy you have available. But you can never get it all the way to c.

 

[andrien]

Time will slow down as it's speed increases(proper time)

 

[SecretSnow]

Why is it that for massless particles, E=pc? How do I prove this? Isn't that for particles with rest mass not equal to zero only?

 

[SecretSnow]

And how can a massless particle have momentum? If relativistic momentum is = gamma*m0*v, then shouldn't it be 0? In this case if the speed is c we have 0/0 which is undefined.

 

[member 11137]

First of all, i need to say that i have read various threads posted by other users in physicsforum who have asked the same question. However, I still don't have a clear picture of why massless particles must travel at c.

Perhaps because, in totally opposition with the academic position, mass itself is not coming from a speed greater than 0 (which is our custumized view point on the earth), but from a speed smaller than c... (which would be a viewpoint from a passenger traveling with the ligth or with the expanding universe)?

Just a poetic proposition.

 

[PeterDonis]

And how can a massless particle have momentum? If relativistic momentum is = gamma*m0*v, then shouldn't it be 0?

Relativistic momentum is only gamma * m0 * v for a particle with nonzero m0. The more fundamental relation is the one mfb posted:

$$E^2 = m_0^2 c^4 + p^2 c^2$$

which works for both massive particles and massless particles. If $m_0 = 0$, the above relation just gives $E = pc$, which is what mfb and jtbell already gave you as the correct formula for the momentum of a photon.

 

[SecretSnow]

Relativistic momentum is only gamma * m0 * v for a particle with nonzero m0. The more fundamental relation is the one mfb posted:

$$E^2 = m_0^2 c^4 + p^2 c^2$$

which works for both massive particles and massless particles. If $m_0 = 0$, the above relation just gives $E = pc$, which is what mfb and jtbell already gave you as the correct formula for the momentum of a photon.

In this case, how do I find the correct value of p? Do I have to use E=hf=pc? Or is there an even more direct way? Thanks a lot! I think I'm getting there already!

 

[mfb]

Correct value of p for what?
E=hf=pc is fine if you know E or f. If not, you need some other way to determine one of those 3 quantities.

 

[PeterDonis]

In this case, how do I find the correct value of p?

If you know E for a massless particle, you automatically know p from E = pc.

 

[UglyNakedGuy]

No.


They do not even accelerate.


$E^2= (mc^2)^2 + (pc)^2$
$E=pc \frac{c}{v}$
For massless particles, E=pc and therefore v=c.

sorry to be an idiot here, how did u put v into the second equation...i mean algebraically

 

[jtbell]

IMO this is easier to show using my version:

$$\frac{pc}{E} = \frac{v}{c}$$

Start with the usual equations for relativistic momentum and energy: $p = \gamma mv$ and $E = \gamma mc^2$ (where $\gamma = 1 / \sqrt{1 - v^2/c^2}$). Substitute these into the left side of the equation above, and cancel whatever you can cancel.

 

[UglyNakedGuy]

IMO this is easier to show using my version:

$$\frac{pc}{E} = \frac{v}{c}$$

Start with the usual equations for relativistic momentum and energy: $p = \gamma mv$ and $E = \gamma mc^2$ (where $\gamma = 1 / \sqrt{1 - v^2/c^2}$). Substitute these into the left side of the equation above, and cancel whatever you can cancel.

I wish I could start my university again with a different course...

thank you :D i got it