# Mass near the speed of light?

1. Jul 11, 2006

### denni89627

mass increased as you approach the speed of light correct? what is the max value- infinity or something else depending on the rest mass? basically what im getting at is how would you calculate the increase in mass based on speed? plus what is speed, after all this is relativity. speed relative to the earth which is also rotating in space? would everythig weigh fractionally less if the earth stopped rotating (without affecting gravity) since you would be in less motion relative to the universe?

2. Jul 11, 2006

### MeJennifer

Well my understanding of it is that mass does not increase in the local frame of reference (this is sometimes called the rest mass of the object), in fact it decreases if the the energy to accelerate comes from the mass in question (due to $$E = Mc^2$$).
However for any other frame of reference that observes a change in distance or direction the mass appears to increase (sometimes called relativistic mass).

3. Jul 11, 2006

### Quaoar

Infinity.

$$m = \gamma m_0$$

where

$$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$

m_0 is the rest mass, v is the velocity of the object, c is the speed of light.

It depends on your frame compared to the object in question. Assume the object is in space, speeding towards the Earth. If you're sitting on the Earth, and the Earth is rotating in a direction so that you are moving towards the speeding object, the object will appear more massive than if the Earth wasn't spinning. If you are moving away from the object due to the Earth's rotation, the object will be less massive.

There is no "relative to the universe." There are no reference points that allow us to fix the universe's position relative to ourselves.

EDIT: Cleared up some confusion.

Last edited: Jul 11, 2006
4. Jul 11, 2006

### MeJennifer

Really?

5. Jul 11, 2006

### Quaoar

I should have specified that the speeding object is moving towards you in the first place...fixed.

6. Jul 12, 2006

### Mute

In my SR course and textbook (Modern Physics, Tippler & ...Llewyn or something?) the view taken is that mass does not increase as you approach the speed of light, but rather the funny business is a property of relativistic momentum. Of course, if I recall everything correctly, if you measure the mass of an object moving at relativistic speeds, the mass you measure will be the rest mass times (1-(v/c)^2)^(-1/2).

7. Jul 12, 2006

### Jorrie

Relativistic mass

Yes, we should refrain from talking about mass that increases with velocity - mass should rather mean 'rest mass' to avoid confusion.

It is coordinate energy and momentum that increase with coordinate velocity, not mass.

8. Jul 12, 2006

### Zeit

Hi,

In this case, should we say that :

"if you measure the duration of a phenomenon moving at relativistic speed, the duration you measure will be the rest duration times (1-(v/c)^2)^(-1/2)"

or

" if you measure the lenght of an object moving at a relativistic speed, the length you measure will be the rest length divided by (1-(v/c)^2)^(-1/2)"

?

Thanks

Last edited: Jul 12, 2006
9. Jul 12, 2006

### MeJennifer

Well I do not see the confusion at all.

I mean would you prefer to use rest elapsed time and rest distance as well because if might confuse?

There is no confusion if we accept that elapsed time, distance and mass are dynamic properties and that their values depend on the frame of reference.

10. Jul 13, 2006

### Jorrie

But there are no things like "rest elapsed time and rest distance". There are coordinates distances and times and there is a thing like rest mass.

Would you assign a dynamic mass to a photon too?

11. Jul 14, 2006

### pmb_phy

Whether mass depends on velocity depends on the definition of "mass." There are two common meanings of the term "mass" as widely used in relativity. One is called "relativistic mass" defined as m = p/v and one is "proper mass" defined as the magnitude of the particle's 4-momentum. Relativistic mass is the velocity dependant one. Proper mass is the velocity independant mass.

Pete

12. Jul 15, 2006

### Mute

Yes, you can view things like that; what I was saying is that my textbook chooses the viewpoint where there is no relativistic mass - it's momentum that changes nonlinearly with speed:

p = [(1-(v/c)²)^-½]mv, where m is the invariant mass

i.e., the factor (1-(v/c)²)^½ acts on the classical (non-relativistic) momentum, not the mass. Thus, mass remains an invariant property of a system, independent of the frame of reference - it's momentum whose value depends on the frame of reference in which it is measured.

Of course, the choice is kind of arbitrary, but perhaps more aesthetically pleasing to some.

13. Jul 15, 2006

### pmb_phy

What's the name and author of your text?
Yes. I'm aware of these things. Thanks. For a detailed description you can read a web page I created on what you're speaking of. See

http://www.geocities.com/physics_world/sr/invariant_mass.htm

Pete

14. Jul 15, 2006

### Mute

The book is "Modern Physics", 4th edition, by Tippler and Llewellyn.

The exact phrasing on the matter is:

"The quantity m(u) [=m/(1-(v/c)²)^½] in equation 2-5 is sometimes called the relativistic mass; however, we shall avoid using the term or a symbol for relativistic mass: in this book, m always refers to the mass measured in the rest frame. In this we are following Einstein's view. In a letter to a colleague in 1948 he wrote:

It is not good to introduce the concept of M=m/(1-(v/c)²)^½] of a body for which no clear definition can be given. It is bnetter to introduce no other mass than "the rest mass" m. Instead of introducing M, it is better to mention the expression for the momentum and energy of a body in motion."
--page 72 in the book

There are a couple of minor errors in the "Invariant mass of a single particle" section: you accidentally write m = γm[0]c², and you state "the magnitude of U, is c². I.e. U·U = c²" when of course |U| = (U·U)^½, so |U| should be c.

15. Jul 15, 2006

### pmb_phy

Actually if one wishes to be "in keeping" with Einstein one must use relativistic mass with v = 0. In general this is not the proper mass of the particle. When there is a gravitational field present gamma does not equal 1 when v = 0. Einstein used this in his relativity text in regards to Mach's Principle and the relationship between the gravitational field and the particle's inertial mass.

Thanks for the corrections to my web page.

Pete

16. May 11, 2011

### chis

A simple answer to the original question using a double engine steam train. You drag a carriage full of coal to burn to go faster and the coal carriage is part of your train, a closed system. The first trip you only use the one engine and hit 100mph and your coal just runs out as you enter the last station. You want to go even faster on the return trip, say up to 150, so you must fire up the second engine, but need an additional coal carriage to feed it to get all the way to you destination. More speed equals more weight in the system and to push on to 180mph you would need another engine and coal carriage......etc.
What ever your mode of transport and energy source to drive it you always end up in a deminishing returns situation when you want to go faster.

Last edited: May 12, 2011