# Relativistic Mass

## Main Question or Discussion Point

From what i heard, my friend told me relatvistic mass has just been recently proven wrong. Is that right????

## Answers and Replies

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misogynisticfeminist said:
From what i heard, my friend told me relatvistic mass has just been recently proven wrong. Is that right????
No this is not right.

How would one prove relativistic mass to be wrong ???

marlon

misogynisticfeminist said:
From what i heard, my friend told me relatvistic mass has just been recently proven wrong. Is that right????
No. Absolultely not. Where did he get that idea from??

Pete

Janus
Staff Emeritus
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Attention:
Tempest99 is just a new user name for Physicsguru. Accordingly, I have deleted all of his posts and any posts referencing them.

Thanks Janus for getting rid of these crappy posts

marlon

Tom Mattson
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misogynisticfeminist said:
From what i heard, my friend told me relatvistic mass has just been recently proven wrong. Is that right????
You can't prove it wrong, because it's a definition. Take the expressions for momentum and total energy in natural units:

$$p=\gamma m_0 v$$
$$E=\gamma m_0$$

If you want to define mass as the (Lorentz-invariant) norm of the 4-momentum, then $m_0$ is that mass. It doesn't change with velocity. On the other hand, there is nothing stopping you from associating $\gamma m_0$ together and defining that product as $m$. This $m$ is called the relativistic mass, and it is a function of velocity.

So, the only way to disprove relativistic mass is to disprove the associative property of real numbers under multiplication (it can't be done).

pmb_phy said:
Where did he get that idea from??
Maybe his name is DW. :rofl:

Tom Mattson said:
On the other hand, there is nothing stopping you from associating $\gamma m_0$ together and defining that product as $m$. This $m$ is called the relativistic mass, and it is a function of velocity.

So, the only way to disprove relativistic mass is to disprove the associative property of real numbers under multiplication (it can't be done).
I'll have to disagree with you on this point. Relativistic mass is defined such that the quantity mv is always conserved in collision (wherein one then defines the product mv as 'momentum'). Had nature been such that this is never measured in the lab then it would follow that the definition is incorrect in that it can't fit into what nature had in mind.
Maybe his name is DW. :rofl:

Pete

dextercioby
Homework Helper
Tom,since "c" is absolute (and NOT BECAUSE IT IS CONVENTIONALLY TAKEN AS "+1"),one should not speak about RELATIVISTIC MASS,but about ENERGY...

Daniel.

Tom Mattson
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pmb_phy said:
I'll have to disagree with you on this point. Relativistic mass is defined such that the quantity mv is always conserved in collision (wherein one then defines the product mv as 'momentum').
Is that not what I said? If you define $\gamma m_0$ as $m$, then $p=mv$ and $E=m$ (in natural units). These quantites are conserved.

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Tom Mattson
Staff Emeritus
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dextercioby said:
Tom,since "c" is absolute (and NOT BECAUSE IT IS CONVENTIONALLY TAKEN AS "+1"),one should not speak about RELATIVISTIC MASS,but about ENERGY...
I'm not sure I get what you are saying. $E$ and $m$ are only identical in natural units. In the SI system, they have different dimensions.

dextercioby
Homework Helper
There's no need to introduce RELATIVISTIC MASS (a rather confusing concept,at least to the unfamiliar with subtleties of SR/GR),when u have already energy...

Daniel.

dextercioby said:
There's no need to introduce RELATIVISTIC MASS (a rather confusing concept,
What pat of it are you confused about?
...at least to the unfamiliar with subtleties of SR/GR),when u have already energy...
Relativistic mass is defined differently than energy. Under certain cercumstances one is proportional to the other, I.e. E = -mc^2. It appers to me that you're asserting that the "m" in m = E/c^2 is identical to the "m" in m = p/v. If so then the assertion is incorrect since, in the general, E/c^2 does not equal m = p/v.

Pete

chroot
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Admin note: Let's not make the same old arguments about relativistic mass all over again. Otherwise, this thread will be locked.

- Warren

dextercioby
Homework Helper
Okay,neglect gravitational potential in that square root.Are u claiming that the "m"-s are different in SR...??

Daniel.

dextercioby
Homework Helper
Okay,Warren,i'll back off.

Daniel.

Tom Mattson said:
I'm not sure I get what you are saying. $E$ and $m$ are only identical in natural units. In the SI system, they have different dimensions.
I was saying that relativistic mass is defined such that mv is a conserved quantity. What you posted implied that the definition cannot be wrong. What I posted was why a definition can be wrong. Its a fine point though and not worth worrying about.

Pete

Tom Mattson
Staff Emeritus
Gold Member
pmb_phy said:
I was saying that relativistic mass is defined such that mv is a conserved quantity. What you posted implied that the definition cannot be wrong. What I posted was why a definition can be wrong. Its a fine point though and not worth worrying about.
I get you. What I should have said was:

Taking SR for granted, relativistic mass cannot be disproven.

Naturally, relativistic mass would be falsified if SR were falsified.

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