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Susskind said that the square of a differential equal zero 
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#1
Apr2814, 06:19 AM

P: 22

Hello,
I watched a lecture by Leonard Susskind, in which he said that a differential is so short that when you square it, you get zero. What exactly could he mean by this? Thank you for your time. Kind regards, Marius 


#2
Apr2814, 07:02 AM

P: 2,969

Perhaps you could post a link to the video and the time inside the video that he said this.
He's probably referring to the size of the number for example if your differential was numerically evaluated to 10^(8) then the square of it would be 10^(16) which is a much smaller number and could be considered to be zero. The effect of this would allow you to drop higher order terms in a power series expansion of a function as an example to get a linear approximation of the function. 


#3
Apr2814, 07:09 AM

PF Gold
P: 374

My guess:
He might mean that the boundary of a boundary is empty. The differential operator he is talking about would be the "exterior derivative". It is this operator that for example allows you to replace a (closed) surface integration by an integration over the volume bounded by this surface. On the basis of this geometric interpretation, you can guess that the "derivative" of a "derivative" is alway zero! (exterior derivative is implied!). The boundary of the volume of a sphere is the surface of the sphere, which is closed. The boundary of the surface of a sphere is empty. Considering an infinitesimal volume in place of a sphere leads you to one special case of the rule dē=0 . http://en.wikipedia.org/wiki/Exterior_derivative http://en.wikipedia.org/wiki/Boundary_(topology) 


#4
Apr2814, 07:57 AM

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PF Gold
P: 12,157

Susskind said that the square of a differential equal zero
(δx)squared will vanish (can be ignored), compared with δx. It's a common way to find how an expression will behave as one of the variables approaches zero. Your "differential" doesn't actually become Zero, it just is infinitely small compared with other things he's dealing with. This link goes through the basics of how differentiation works and, right at the end, show how you ignore terms with dx in. 


#5
Apr2814, 09:01 AM

Mentor
P: 15,164

A link to the video would be nice.
I suspect it is related to his book The Theoretical Minimum: What You Need to Know to Start Doing Physics (coauthored with George Hrabovsky). It apparently uses "physics math", where one ignores the Δx^{2} terms (and higher) in calculating derivatives. It's much easier than the more rigorous epsilon delta approach to a limit. 


#6
Apr2814, 11:05 AM

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P: 1,948

That's sometimes called an infinitesimal of higher order. Δx^{2} is and infinitesimal of higher order and can be dropped from calculations. The point is that there will be an infinitesimal of first order somewhere in the denominator of the expression that will cancel infinitesimals of first order in the numerator (or/and vice  verse). But infinitesimals of higher order do not get cancelled out and will eventually drop out when the limit is taken. That limit is often implicit but one needs to keep in mind that it's there.



#7
Apr2814, 12:05 PM

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#8
Apr2814, 02:59 PM

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#9
Apr2814, 03:32 PM

PF Gold
P: 1,503

Sounds to me a lot like what is commonly done in an order of magnitude analysis when linearizing an equation or making other similar approximations to make a problem more tractable.



#10
Apr2814, 05:32 PM

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#11
Apr2814, 06:01 PM

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#12
Apr2814, 06:56 PM

P: 110

wow Susskind is amazing, I was wondering too how small can you reduce a unit basis vector, maybe to the size of a differential?
I guess very few people have the talents to become theorists? 


#13
Apr2814, 08:28 PM

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#14
Apr2814, 08:39 PM

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well say you could create a vector particle whose integral from 0 to infinity would be 1 unit vector :)
this seems to be a powerful technique for instance to discretize a space.... 


#15
Apr2814, 10:54 PM

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I'm not sure what a 'vector particle' is. It's not clear how vectors could be used to discretize space, since space is not a vector quantity.



#16
Apr2914, 09:49 AM

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