# Can you speak English for a lowly S/W Dev?

1. Feb 7, 2004

### treat2

Can you speak English for a lowly S/W Developer, please... I have always had an interest in physics, and have done some reading, on it, and will do much more. I'm not a mathematician, nor a Physicist, but I can catch on to ANYTHING well explained!

Here is the preliminary to the question I have concerning the increase in mass, as velocity increases. BUT first, I want to point to some Web Pages that confirm that a photon DOES have mass, as I'm not sure if it is part of my confusion or not.

1) "What is the Mass of a Photon?" See:
http://math.ucr.edu/home/baez/physi...hoton_mass.html [Broken]

2) For a discussion of it, in another String Theory Forum, See:
http://superstringtheory.com/forum/...sages10/23.html [Broken]
(Many other Web Sites can be found stating the Mass of a Photon).

If I recall correcly, I read on one Web Site that the mass of a photon is estimated at 10^-48th of the mass of an electron.

Now for my questions... THIS MUST BE VERY BASIC STUFF FOR YOU FOLKS

1) What is meant by "infinite mass", when speaking of the mass of an object traveling at light speed.

2) Since a photon is traveling at light speed, why does it not have infinite mass; if not infinite mass, then at least the mass of a planet?

3) Along with infinite mass should come an infinite gravitational pull. So, why wouldn't an object with "infinite mass", have "infinite
gravity", which leads to two questions:
3a) What is infinite gravity?
3b) Why wouldn't the gravitational pull of the photon (which is supposed to have "infinite mass"), cause the entire planet to be swallowed into it, just like a Black Hole would do???

Again, I appologize for my ignorance, but these are the words used in physics books I've read, and I'm just following what I think is supposed to happen for a particle with mass travelling at light speed.
Thanks for your patience. I look forward to reading a reply that doesn't flame my ignorance. Thanks again.

Last edited by a moderator: May 1, 2017
2. Feb 7, 2004

Staff Emeritus
I'll just do this bit.

If I recall correcly, I read on one Web Site that the mass of a photon is estimated at 10^-48th of the mass of an electron.

That is their upper limit considering unpreventable experimental error, i.e. if the photon had any mass it couldn't be bigger than that. There are very strong theoretical reasons to suppose the photon has no mass:

1) If the photon had mass then electromagnetic radiation would have a longitudunal component, and to the limits of experimental error, it doesn't (this argument goes back to Isaac Newton, no less, and it is actually the basis for the measurement you mentioned).

2) The photon interactions in the relativistic quantum electrodynamics would come out wrong, and they wouldn't be able to calculate things like the Lamb shift and the anomalous magnetic moment of the electron as accurately as they do. This accuracy is the best that any physics theory has ever generated; calculation matches experiment to six decimal places.

So physicsits feel pretty sure the photon is massless, and experiment doesn't contradict them whatever some crank websites may assert. Every time the experimenters are able to increase their accuracy, that upper limit goes down, and experimentalists will tell you, there is NO EXPERIMENTAL EVIDENCE FOR PHOTON MASS.

3. Feb 7, 2004

### pmb_phy

You're confusing the two different uses of the term "mass." That link you provided explains this difference. An increase in speed yielding an increase in mass refers to what is sometimes known as "relativistic mass." This mass can be thought of as the ratio of the magnitude of a particle's momentum to its speed, i.e. m = p/v. When it is said that light has zero mass it refers to "proper mass" aka "rest mass."

Point by point
It means that, given a particle with a finite proper mass, m0, as the particle's speed approaches the speed of light its momentum increases without bound. This follows from the relations

$$\mathbf{p} = m\mathbf{v}$$

$$m = \gamma m_{0} = \frac {m_{0}} { \sqrt{1-v^{2}/c^{2}} }$$

Since both the momentum and the velocity of light are finite then so is the ratio p/c = finite.
That is true. If you're in a gravitational field in which you measure a finite gravitational acceleration of a falling particle then if you change to a frame of reference which is moving with respect to that frame then the magnitude of the gravitational acceleration as measured in this "moving" frame will increase. For example: Suppose you have a large sheet of matter which generates a uniform gravitational field. Assuming the pressure of the matter is small compared to the mass density then the gravitational field in the rest frame of the matter will give rise to a gravitational acceleration which is proportional to the surface mass density (mass per unit area) of the sheet. Let this acceleration be g0. Now transform to a frame of reference moving with respect to the sheet. There will be an increase in the relativistic mass of each portion of the sheet and a decrease in the volume of each portion. There will therefore be a gamma factor for each. In the moving frame the instaneous gravitational acceleration (asusming falling particle has zero speed) will then be g where

$$g = \gamma^{2} g_{0} = \frac {g_{0}} {1-v^{2}/c^{2} }$$

As the speed of the sheet goes to c then g goes to infinity.

That should take care of the rest of the questions.

I hope that helps.

Last edited: Feb 7, 2004
4. Feb 7, 2004

### HallsofIvy

Well, I would interpret it as saying that any object with mass can't travel at light speed!

Yes, any object with non-zero "rest-mass" would have infinite mass when traveling at light speed- which is why it can't! A photon, however, has no rest-mass and so has no mass at all even though it is moving at light speed.

5. Feb 7, 2004

### GRQC

No. This is incorrect. Mass is a relativistic invariant quantity. The biggest confusion I have seen on this board in the past 2 days is the distinction between mass (magnitude of a 4-dimensional vector) and energy (time component of that vector).

The energy required to accelerate the object will grow to infinity as the velocity $\rightarrow c$, but the mass stays the same. That is, energy and momentum are relative quantities, but mass is not.

A very good exposition on this confusion can be found in:

Taylor and Wheeler, Spacetime Physics: Introduction to Special Relativity (2nd ed).

See pp. 246-251, "Dialog: Use and Abuse of the Concept of Mass"

All ye who covet the phrase "relativistic mass" should read and heed...

6. Feb 8, 2004

### pmb_phy

To be precise one would say that if a particle has a non-zero proper mass then it can't be accelerated to the speed of light from a speed which is greater than or less than the speed of light. However if a partilce is "born" traveling at a speed greater than or less than the speed of light then it will always travel at a speed greater than or less than the speed of light respectively.

Particles are classified according to this scenario

Tardyon - Speed always less than c
Luxon - Speed always equal to c
Tachyon - Speed always greater than c

where "c" refers to the speed of light as measured locally and the speed mentioned above refers to the locally measured speed.

7. Feb 8, 2004

### meddyn

Treat2, I also develop software for the mundane world of invasive cardiology and thoracic surgery measuring and interpreting biological anomalies. Reading the forum over time has led to a firm understanding that questions agreeing with ordinary logic are often illogical ("relative" to an ordinary viewpoint) mathematically. It seems quite simple (relative to ordinary logic) that a galaxy traveling away from us at 100 miles per second is emitting light in the opposite direction at C + 100 miles per second and emitting light back at us at C - 100 miles per second.

The answer, of course, depends on where the viewpoint is and some really impressive formulas that say it is and is not true, depending. And to embelish your questions, if the photon has no mass then "ordinary logic" ask how can light be gravitationally "lensed" around a large mass body. Doesn't gravity act on mass? I think the reply there really does create some interest (and good formulas too).

One thing I was very impressed with recently was a "Law" I heard in a course on Emergency Medicine. "Do not rely on your logic in treatment; rely on your training and education. The injury/disease vs. treatment that is appropriate may not be logical at your level". So, at my level, I just continue to admire and be impressed with this forum while learning some neat stuff.

8. Feb 8, 2004

### pmb_phy

It might seem that way but that has never been observed. People made the assumption for a long time that the speed of light was a function of the velocity of the source emitting the light. However all experimental efforts to verify this assumption met with failure. For this and other reasons in 1905 Einstein postulated that the speed of light in vacuum is indpendant of the motion of the source. As such the speed of the source does not play a role in the speed of light emitted from the source. To be exact the postulate was phrased in his paper as
Predictions based on that assumption have been borne out in all experiments executed to date.
This confusion arises when once fails to distinguish the difference between mass, m, (relativistic mass) and proper mass. m0.

The mass (i.e. relativisitc mass), m, of light is the ratio of the momentum of light to the speed of light. The proper mass of the photon is zero.

Think of proper mass as you would proper time and mass as you would time.

Last edited: Feb 8, 2004
9. Feb 8, 2004

Get off your bloody pedastal, you know what we mean, we are referring to realtivistic mass. Bottom line: Kinectic energy increases the effective mass, the mass that one must deal with in equations such as $$F=ma$$ with $$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$. You insisting on saying that $$F=\gamma*m_0a$$ with $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} [\tex] is exactly equivalent. IT MAKES ABSOLUTELY NO DIFFERENCE. So just get over it. 10. Feb 8, 2004 ### GRQC It makes a very big difference. There is no such thing as relativistic mass. If you're going to learn proper relativity theory, you must know the difference. There is a very important difference between mass and energy-momentum that must be acknowledged. In fact, I would avoid $\gamma m_0 a$ for relativistic force, and use $\frac{dp}{dt}$ instead. This removes any confusion about "mass" from the picture, and introduces the more appropriate relativistic variable, 3-momentum. You cannot write $m = \gamma m_0$ because rest mass is a relativistic invariant! Have you ever see relativity texts talk about "relativistic proper time", $\tau = \gamma \tau_0$? I'm afraid I'm not to one who has to get over it. 11. Feb 8, 2004 ### pmb_phy Yes. That is 100% correct. What do you mean by this? The definition of force is not f = ma. It's f = dp/dt where p = mv. It's this later expression which defines mass, not the first. That is a relationship between the transverse component of force, the transverse acceleration and the transverse mass = relativistic mass. In that expression the F is transverse component of force and so it does not hold when F is the total force. Last edited: Feb 8, 2004 12. Feb 8, 2004 ### franznietzsche $$p = mv m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} F = \frac{dp}{dt} = \frac{dm}{dt}v + m\frac{dv}{dt}$$ now assuming mass is invariant, this reduces to F = ma. For our case where mass increases only with velocity, and not independently of time we get: $$F = \left(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\right)\frac{dv}{dt}$$ Which is relativistic mass times acceleration. Ergo, F = ma. That is what i meant. Using $$\frac{dp}{dt} [tex] is unnecessary except in the cases where mass varies dependent on time, and those cases are compartively rare, at least that i know of. Also it is far easier to experimentally measure rest mass, then apply the relevant Lorentz transformation than it is to measure the change in momentum. F=ma is easier to use for experimental purposes. The use of that equation implies that rest mass is invariant. However when dealing with forces acting on an object, rest mass is useless(except in the exception of stationary objects, but since a force is acting on them they will not long remain stationary, we must know the the relativistic mass. Edit: Why is the tex not working? and the quote for that matter? Its been working on/and off all day for me.... Last edited: Feb 8, 2004 13. Feb 8, 2004 ### GRQC Last edited: Feb 8, 2004 14. Feb 8, 2004 ### pmb_phy First off please note that you're using the term "invariant" to mean "does not depend on time". The term "invariant" in relativity is usually taken to mean "does not change upon a change in coordinates." The only way that dm/dt can be zero is for a = 0. Do this calculation out explicitly and you'll see what I mean and note that v = v(t). Note that if m is a function of v then for an accelerating particle v = v(t) and therefore m = m(t). So this is not a special case but is the general case. I.e. the mass of a particle changes when there is a force acting on it. Actually for subatomic particles its just as easy to measure relativisitc mass as it is to measure rest mass. Probably easier as a matter of fact. Take the case of an electron as an example. Shoot an electron into a uniform magnetic field perpendicular to the field lines and measure the radius of curvature of the path and the cyclotron frequency and with the value of the magnetic field strength you can calculate the value of the relativistic mass. See http://www.geocities.com/physics_world/sr/cyclotron.htm That is an incorrect statment. The term relativistic mass is the name given to the quantity m = m0/sqrt[1-(v/c)^2]. Proper mass is the invariant quantity in that relation, not relativistic mass (although there is a context in which one can say that relativistic mass is a scalar). In relativity the qualifier "relativistic" does not mean "invariant" it means "precise at all speeds less than the speed of light." That is the reason that [tex]\mathbf{p} = \frac {m_{0}\mathbf{v}} { \sqrt{1-v^{2}/c^{2} } }$$ is refered to as relativistic momentum (e.g. Ohanian 2001). You have the slash going the wrong way when you end the "tex". i.e. it should read "$$" but you're writing it as "[\tex]"

Last edited: Feb 8, 2004
15. Feb 8, 2004

### meddyn

So, Treat2, after carefully reviewing and testing the replys I worked out a formula based on relativity:

U = (MC^2) - E

Let U = Understanding

16. Feb 8, 2004

### franznietzsche

refreshing humor indeed.

17. Feb 9, 2004

### DW

Actually the site says the opposite of what you claim. In specific
"Does light have mass?
The short answer is "no", but it is a qualified "no" because there are odd ways of interpreting the question which could justify the answer 'yes'."
http://tinyurl.com/ywvww

The only other of the "pages" you list as far as I can tell is
http://tinyurl.com/22r8d
which is not a verification, but a layman discussion board.

And they are wrong if aren't listing a null result.

No that is the maximum mass that it could have been due to limitations on the experiments precision.

Mass does not change with speed. Things that travel at the speed of light don't have infinite mass. They have zero mass.

Why would it? Things that travel at the speed of light have zero mass.

Nothing has infinite mass so it wouldn't matter, but even so relativistic gravitation is not given by Newton's law of gravitation.

Something that doesn't exist.

Its not supposed to have any such thing. Its supposed to have zero mass.

Since it has zero mass and its gravitation isn't given by Newton's law of gravitation, why should that be so?

At the least you have provided an excellent example of why in the modern relativistic terminology the the concept of "relativistic mass" has been done away. I recommend getting better books. If it must be at the undergrad level then I recommend any recent text by Serway that has a chapter on at least special relativity.

Last edited by a moderator: May 1, 2017
18. Feb 9, 2004

### DW

Re: Re: Can you speak English for a lowly S/W Dev?

This is a good example of why you should stop using this obsolete concept.

Correction, the mass m is not the relativistic mass M = p/v.

It means just exactly what it said, zero mass.

Should be written just m as the zero subscript is meaningless. The mass is invariant. It is the same for every frame, not just the value of the mass for the rest frame.

Should be $$P^i = mU^i = \gamma mu^i$$.

Should be $$E = \gamma mc^2 = \frac {mc^{2}} { \sqrt{1-v^{2}/c^{2}} }$$

And by that line of reasoning if you change to a frame of reference which is moving with respect to the second and just happen to do get back to the original motion state by that boost it must increase yet again! Every frame is in motion with respect to some frame, even your original frame.

(snipped the rest as its irrelevent to his question at hand)

Last edited: Feb 10, 2004
19. Feb 9, 2004

### DW

Like you are leading people to do right now. It should be
...between mass $$m$$, and (relativistic mass) $$M = \gamma m$$.
And you shouldn't even be using the latter.

Correction, ...relativisitc mass), M....

Correction - The mass of the photon is zero.

Argument by analogy does not constitute a logical argument. As I told you befor, the relativity of spacetime coordinates is not analogous to the invariance of the physics.

20. Feb 9, 2004

### DW

This is why you need to stop referring to relativistic mass and using it at all. Your equation is just plain WRONG.