Why Does Approaching the Speed of Light Affect Mass and Time?

[semidevil]

so I"m just reading about physics for fun these days, and I"m still trying to understand the concept of relativity.

So why does mass get higher and time for me go slower as I approach the speed of light? I know this has to do with the concept of relativity and frame of reference, I"m trying to grasp the basic concept of it.

anyone care to explain to me in simpler terms?

thanks

 

[baryon]

I hope someone posts an answer to that question soon. How does energy get converted to mass? I've never quite understood myself. If I come across anything I'll post it.

 

[Oxymoron]

Basically, these things happen because of the math.

Here are some hints that should lead you to your answers:

Posted by Semidevil:

So why does mass get higher...as I approach the speed of light?

F=ma[/itex]<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Posted by Semidevil:<br /> <br /> So why does ... time for me go slower as I approach the speed of light? </div> </div> </blockquote><br /> \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}<br /> <br /> Let me know if you need some more explanation.

 

[DaveC426913]

Basically, these things happen because of the math.

Here are some hints:
F=ma[/itex]<br />

<br /> <br /> This is a pretty bizarre hint, since it&#039;s the basis of classical Newtonian mechanics and has nothing to do with relativity.

 

[Oxymoron]

Posted by Baryon:

How does energy get converted to mass? I've never quite understood myself.

The famous E=mc^2 equation tells us that mass (or matter) is energy. The way I like to think of it is this: Matter (which has mass, always!) is what you get when you "squeeze" a whole lot of energy into one place - and I mean a lot of energy. Doing the opposite (that is, breaking the matter up as opposed to putting it together) "spreads" the energy out again.

Notice that E=mc^2 implies that a body at rest has energy which contradicts the Newtonian view of motionless bodies. This leads me to mentioning the fact that you must understand the difference between rest mass and invariant mass. This leads you to the following Special Relativistic form of E=mc^2:

E = \frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}

 

[Oxymoron]

This is a pretty bizarre hint, since it's the basis of classical Newtonian mechanics and has nothing to do with relativity.

Yeah, maybe it was too cryptic.

Obviously F=ma or better yet: F=\frac{d(Mv)}{dt} will not work in SR because Mass is not considered constant. Instead one must correct with the Lorentz factor (my second hint) which you can read about here:

http://en.wikipedia.org/wiki/Relativistic_mass"

 

[baryon]

The famous E=mc^2 equation tells us that mass (or matter) is energy. The way I like to think of it is this: Matter (which has mass, always!) is what you get when you "squeeze" a whole lot of energy into one place - and I mean a lot of energy. Doing the opposite (that is, breaking the matter up as opposed to putting it together) "spreads" the energy out again.

Notice that E=mc^2 implies that a body at rest has energy which contradicts the Newtonian view of motionless bodies. This leads me to mentioning the fact that you must understand the difference between rest mass and invariant mass. This leads you to the following Special Relativistic form of E=mc^2:

E = \frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}

So the increase in energy is translated to an increase in mass. Does that mean that the space that the object is traveling in becomes more curved?

 

[pervect]

So the increase in energy is translated to an increase in mass. Does that mean that the space that the object is traveling in becomes more curved?

Note that while the relativistic mass of an object increases, its invariant mass stays the same.

As far as curvature goes, the curvature tensor itself is usually considered to be a coordinate-independent object, which exists independently of any particular coordinate system. So the object is not considered to be a "different" object when viewed by a moving observer or a stationary observer, it's considered to be the "same" object. This means that the curvature, considered as a tensor, is considered to be the same regardless of velocity.

The curvature tensor at a location in space is not just one number - it consists of 4x4x4x4 = 256 numbers, many of which, however, are constrained to be the same because of symmetry.

However, components of the curvature tensor will vary depending on one's coordinate system. I.e. the values of these 256 numbers will change when one changes one's velocity. They change in a "standard" manner - this is in fact the definition of a tensor quantity, that its components will change in a "standard" manner when one changes coordinates.

 

[michael879]

mass doesn't actually increase as you approach the speed of light. Its pretty misleading the way people say it. What increases is relativistic mass which is rly nothing like rest mass. Relativistic mass is a vector (I think i remember seeing some picture of an ellipsoid) and doesn't really exist. its gamma * mass (where gamma is\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}) and in my opinion is just used to make relativistic equations look like their Newtonian counter-parts. For example, instead of F = \gamma ma you can use F = m_ra (btw that example might be completely wrong, don't really remember if there is a relativistic force equation but it still shows my point).