Why doesn't a photon's mass increase to infinity?

[jbriggs444]

I did not say anyone was disagreeing with it. I was clarifying the confusing way I had written my earlier post.
Where did the bullet come in?

You had surmised that it was impossible to measure a photon's energy and had proceeded to reason about objects on tables or in cars. The apparent thrust of this was that we cannot measure the kinetic energy of an object on a table or of a photon.

Yet we can measure a bullet's energy.

 

[Francis Ward]

I was not aware I had surmised as such and mentioned no tables. The apparent thrust was not that either. Probably best to quit now as the discussion is not a discussion and is fruitless

 

[Herman Trivilino]

they do not [...] experience time. Which is precisely what I stated in the first place. Elementary SR.

You didn't say it precisely. To say that a massless particle doesn't experience time is not a precise way of expressing this idea.

The passage of time is not defined because any two events for which a massless particle is present are separated by an interval of zero. That is, the separation is lightlike. Proper time is the interval length, but only for events that have a timelike separation. If the separation between two events is lightlike or spacelike there is no way of assigning a proper time elapsing between the events. The very concept of proper time doesn't exist, which is quite different from saying that the elapsed proper time has a value of zero. In the latter case the implication is that the concept exists, after all, you can't assign a value (of zero or anything else) to a quantity that is undefined.

 

[PeterDonis]

I also checked the definition of four momentum in Rindler, Essential Relatvity. This is also mixes momentum, mass and energy in its definition. I made some margin notes years ago when I first noticed the inconsistency. There is a simple test for consistency in physics, that of the units. For a four vector to have a modulus, which it must, all components have to have the same units. The forth term in the four momentum is mc, not m. This may be a simple matter of missing universal constants out of expressions, a habit that oft times creates problems.

No, it's a matter of understanding what the constant $c$ actually is in relativity. It's a unit conversion factor. You can use units where $c = 1$ and then the whole issue you describe goes away; all components of 4-momentum have units of mass. Or you can use momentum units, which means you assume a value fori $c$ that is not $1$ (such as the current SI unit value), and insert a factor of $c$ as appropriate in the components of 4-momentum; or you can use energy units, which means in some places you're inserting a factor of $c^2$ (and in others you're inserting a factor of $c$). None of which changes the physics in the least. And none of which changes the point that @Ibix was making, which he stated again in post #56, and which is perfectly valid.

 

[Orodruin]

There is a simple test for consistency in physics, that of the units. For a four vector to have a modulus, which it must, all components have to have the same units. The forth term in the four momentum is mc, not m. This may be a simple matter of missing universal constants out of expressions, a habit that oft times creates problems.

This would be true if you were using a system of units where length and time dimensions were different. Rindler is not doing this and it is a very common thing to do in relativity to use the same units for length and time by using units where, by definition $c = 1$, just as it is common in quantum physics to use units where $\hbar = 1$. The typical thing to do in high-energy physics is to use units where $\hbar = c = 1$ and measure lengths and times in inverse units of energy. Understanding the system of units used is as fundamental as dimensional analysis.

In relativity, it is natural to use the same units for time and length, after all the manifold is spacetime and using the same units for length and time just assures that all your coordinates have the same physical dimension. Note that all velocities are dimensionless in those units.

If you will, this is equivalent to considering the spatial coordinates to be $x/c$ rather than $x$ (or the time coordinate to be $x^0 = ct$ rather than $t$).