infinite enery

[harsh]

I have read that when a body comes closeto the speed of light it undergoes a change in mass and its mass becomes infinite , well how is it possible wont it require infinite energy in order to attain infinite mass??...Please do correct me if a wrong awaiting your replies.

thank you

[fuhrerkeebs]

The body attains infinite mass only when it reaches the speed of light, which is impossible for the reasons you've stated.

[harsh]

ohh ok so only once it reaches speed of light it gains infinite mass......

but if we say that light is energy given out by conversion of mass then should light have mass , weight , and is light considered to be matter??

[YT2095]

light (the photon) has no mass, that`s why it can travel at such speeds without violating any laws

[harsh]

i am a lil confused....i order for anything to move it will require energy and wont light(photon) challenge the basic formulas of e=mgh , e=1/2mv^2??....it may sound stupid but..............

[Aeschylus]

1) minor nitpick: mass, by the defintion of mass, does not increase with speed.

 

2) accelerating to the speed of light is impossible. For example: this cannot be acheived with a finite accelartion.

 

3) the energy of a photon is related to it's frequency/wavelength, formulas like KE = 1/2 mv^2 will not give you the KE of a massless particle like a photon.

[Martin]

i am a lil confused....i order for anything to move it will require energy and wont light(photon) challenge the basic formulas of e=mgh , e=1/2mv^2??....it may sound stupid but..............

 

I agree with what Aeschylus just said

 

it's fun to use the LaTex math notation tool that blike or someone recently turned on at SFN

so I will write down the actual basic KE formula for something with positive (rest) mass m traveling at speed v.

 

it actually is not

 

[math]KE = \frac{1}{2}m v^2[/math]

 

that's actually the wrong formula, but for low speeds like what we are used to it is a DARN GOOD approximation of the correct formula. The correct formula is:

 

[math]KE = (\frac{1}{\sqrt{1 - (v/c)^2}}-1)m c^2[/math]

 

Mother Nature, in her infinite kindness (violins playing here) does not require us to use the more complicated formula for the KE except for

particles in physics toys and other things whizzing at substantial fractions of the speed of light. We never have to use the correct complicated formula in everyday life----because for any speed v less than one percent of c the good old simple formula gives a very good approximation. You almost cant tell the difference!

-----------------------

the root meaning of mass is inertia

nowadays most physicists take an objects mass to mean its

inertia at rest because there are some problems defining

the inertia of a moving object (it is different whether you push sideways or in line with the motion). so when they say mass they mean what is sometimes also called "rest mass".

 

the trouble with saying rest mass all the time is that it suggests that there should be some other kind! also it is a bother and saying just mass is easier.

 

there is a minority of physicists who get furious when one defines mass as rest mass because they want to define mass as "relativistic mass" dependent on the speed. so long sterile arguments can ensue merely about words.

 

Indeed an object's resistence to further acceleration gets bigger as it speeds up, so it becomes harder and harder to accelerate it further.

this is maybe the best way to think about the speed limit. You cant reach c because the particle gets incredibly sluggish as it nears c, and it becomes harder and harder to goad it to go any faster

 

the REST MASS stays the same thru all of this because it is the inertia you would measure if you stopped the particle and measured its inertia at rest.

 

In what Aeschy said, and in the formula I gave, the positive mass m is the rest mass. It avoids some possible confusion to stick with that.

[Janus]

1) minor nitpick: mass' date=' by the defintion of mass, does not increase with speed.

 

2) accelerating to the speed of light is impossible. For example: this cannot be acheived with a finite accelartion.

 

3) the energy of a photon is related to it's frequency/wavelength, formulas like KE = 1/2 mv^2 will not give you the KE of a massless particle like a photon.[/quote']

 

To further nitpick

 

[math]KE=\frac {mv^{2}}{2}[/math]

does not give you the KE of a massive particle, though it comes close with values of v that are small compared to c.

 

The total energy of a massive particle is

 

[math]E = \frac{ mc^{2}}{\sqrt{1-\frac{v^2}{c^2} }}[/math]

 

converted to a series, you get:

 

[math]E = mc^2 + \frac{mv^2}{2} + \frac{3mv^4}{8c^2} +. . . .[/math]

 

The first term is the expression of the energy equivalence of the the rest mass of the particle, so we remove it to leave the KE:

 

[math]KE = \frac{mv^2}{2} + \frac{3mv^4}{8c^2} +. . . .[/math]

 

The first term is now the Newtonian formula for KE, and the rest of the terms don't become appreciable until v becomes some fair fraction of c, so we generally ignore them in everyday life.

 

Reverting this formula back to its original form we get:

 

[math]KE = mc^2 \left(\frac{1}{\sqrt{1-\frac{v^2}{c^2} }} -1\right)[/math]

[Aeschylus]

It's true, the KE of any partilce is not given by 1/2 mv^2 except in the degenrate case. But if were going to triple nitpick, the last formula doesn't give you the energy of a massless particle as it's undefined for v = c.

[Martin]

It's true, the KE of any partilce is not given by 1/2 mv^2 except in the degenrate case. But if were going to triple nitpick, the last formula doesn't give you the energy of a massless particle as it's undefined for v = c.

 

I agree again with Aeschylus! In my post I gave the same KE formula as Janus, and specified that the mass m had to be positive.

zero mass is a separate case---you dont have a formula for KE involving speed and the speed can only be c

 

maybe someone should write the even more basic "energy momentum relation". would anyone care to do the honors?

 

I am wondering what the original poster really needs to be told, at this point

 

actually Janus may have left out a square root sign---but I believe we were both thinking of the same formula

[fuhrerkeebs]

To further nitpick

 

KE= mv²/2 does not give you the KE of a massive particle' date=' though it comes close with values of v that are small compared to c.

 

The total energy of a massive particle is

 

E = mc²/(1-v²/c²)

 

converted to a series, you get:

 

E = mc² + mv²/2 + 3mv^4/8c²+. . . .

 

The first term is the expression of the energy equivalence of the the rest mass of the particle, so we remove it to leave the KE:

 

KE = mv²/2 + 3mv^4/8c²+. . . .

 

The first term is now the Newtonian formula for KE, and the rest of the terms don't become appreciable until v becomes some fair fraction of c, so we generally ignore them in everyday life.

 

Reverting this formula back to its original form we get:

 

KE = mc²(1/(1-v²/c²) -1)[/quote']

 

 

Since everyone else is doing it, I thought I'd nitpick a bit too. Energy expanded in terms of taylor series is actually E=mc^2+(1/2)mv^2-3mv^4/8c^2... (I don't know how to do superscripts)

[Martin]

hello keebs

You are cordially invited to try this for superscripts

 

put the words "sup" and "/sup" in brackets

 

 

if you want to see how somebody writes something, press "quote" under their post as if you were going to quote and reply-----this doesnt obligate you to reply, you just get to see what they typed

 

E = m c²

 

I will damage this by replacing the u in sup with an asterisk

just so it will explicitly show what I typed

 

E = m c[s*p]2[/s*p]

[fuhrerkeebs]

Nice...thanks!

[Martin]

You are heartily welcome!

Anytime you want to try using LaTex, just do the same "quote" trick on any post with a sample of it

My post about 7 posts back has a sample

 

you will see that the important thing is to put "math" and "/math"

in brackets at the beginning and at the end of what you write

 

[m*th]E = m c^2[/m*th]

 

this, with * replaced by letter a, will make

 

[math]E = m c^2[/math]

[Martin]

BTW keebs, Janus is an excellent fellow

I believe I have seen his posts elsewhere and they are quite good.

I havent seen him around here at SFN much so far but if he decides

to stick around it will be a bit of luck for us.

At the other place he had this sig which said

let's see, it was a quote from the philosopher and mathematician Betrand Russell, it said IIRC

"The whole trouble with the world is that fools and fanatics are always so certain of themselves, and that wiser people are so full of doubts."

 

Lot of truth there. Even greens (who I sympathize with for loving the planet and nature etc.) are beginning to bug me because of this passionate selfrighteous certainty they get. they are as bad as neocons. Well this is off topic.

[Janus]

 

actually Janus may have left out a square root sign---but I believe we were both thinking of the same formula

 

Yep, I did. I'll go back and edit. Thanks.

[Thales]

I have a question. If lorentz contraction dictats that distance shrinks at 1/gamma, then wouldn't a photon effectively be travelling zero distance as gamma approaches infinity, or does it asymptote?

 

If as photon travels a very tiny distance, in its reference frame, wouldn't it follow that this distance could be matched to its wavelength (?)

[Aeschylus]

I have a question. If lorentz contraction dictats that distance shrinks at 1/gamma' date=' then wouldn't a photon effectively be travelling zero distance as gamma approaches infinity, or does it asymptote?

 

If as photon travels a very tiny distance, in its reference frame, wouldn't it follow that this distance could be matched to its wavelength (?)[/quote']

 

No, I've tried to impress this point sevral times photons don't have refernce frames, so the question doesn't have any meaning in special relativity.

[Thales]

Cheers Aeschylus.

[unknow force]

Hello, an question, if two bodies at travel near of the light speed, with huge mass, there are an gravitational atraction inbetween??, since the increased mass is relative (for an observer), but for the bodies there are not relative speed????